Question

If a, b, c are in continued proportion, then prove that a/a+2b=a-2b/a-4c

Answer 1

Step-by-step explanation:

As per the question,

a, b, c are in continued proportion that means,

$\frac{a}{b} = \frac{b}{c} =\frac{c}{a}$

This implies that:

$b^{2} =ac$

$c^{2} = ab$

$a^{2} = bc$

Given equation:

$\frac{a}{a+2b}=\frac{a-2b}{a-4c}$

Consider LHS of this equation, we get

$\frac{a}{a+2b}$

On multiplying numerator and denominator by (a - 2b), we get

$\frac{a(a-2b)}{(a+2b)(a-2b)}$

$\frac{a(a-2b)}{(a^{2}-4b^{2})}$

Put $b^{2} =ac$

$\frac{a(a-2b)}{(a^{2}-4ac)}$

$\frac{a-2b}{a-4c}$

Which is equal to RHS

Therefore,

LHS = RHS

Hence, proved.

Answer 2

Step-by-step explanation:

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